Problem: Let $f(x)=x^4+14x^3+52x^2+56x+16$.  Let $z_1,z_2,z_3,z_4$ be the four roots of $f$.  Find the smallest possible value of $|z_{a}z_{b}+z_{c}z_{d}|$ where $\{a,b,c,d\}=\{1,2,3,4\}$.
Solution: Note that
\[\frac{1}{16}f(2x)=x^4+7x^3+13x^2+7x+1.\]Because the coefficients of this polynomial are symmetric, if $r$ is a root of $f(x)$ then $\frac{4}{r}$ is as well. Further, $f(-1)=-1$ and $f(-2)=16$ so $f(x)$ has two distinct roots on $(-2,0)$ and two more roots on $(-\infty,-2)$. Now, if $\sigma$ is a permutation of $\{1,2,3,4\}$:
\[|z_{\sigma(1)} z_{\sigma(2)} + z_{\sigma(3)} z_{\sigma(4)}| \le \frac{1}{2} (z_{\sigma(1)} z_{\sigma(2)} + z_{\sigma(3)} z_{\sigma(4)} + z_{\sigma(4)} z_{\sigma(3)} + z_{\sigma(2)}z_{\sigma(1)}).\]Let the roots be ordered $z_1 \le z_2 \le z_3 \le z_4$, then by rearrangement the last expression is at least:
\[\frac{1}{2}(z_1z_4+z_2z_3+z_3z_2+z_4z_1).\]Since the roots come in pairs $z_1z_4=z_2z_3=4$, our expression is minimized when $\sigma(1)=1,\sigma(2)=4,\sigma(3)=3,\sigma(4)=2$ and its minimum value is $\boxed{8}$.